1. ## sequence

Prove the following: Let $\displaystyle (a_n)$ and $\displaystyle (b_n)$ be sequences of real numbers that converge to real numbers $\displaystyle a$ and $\displaystyle b$ respectively. If $\displaystyle a_n \leq b_n$ for all $\displaystyle n \in \mathbb{N}$, then $\displaystyle a \leq b$.

Proof: For each positive real number $\displaystyle \epsilon$ there exists a $\displaystyle N_1 \in \mathbb{N}$ so that whenever $\displaystyle n > N_1$, $\displaystyle |a_n-a| < \epsilon$. Also for each positive real number $\displaystyle \epsilon$ there exists a $\displaystyle N_2 \in \mathbb{N}$ so that whenever $\displaystyle n > N_2$, $\displaystyle |b_n-b| < \epsilon$. So $\displaystyle a- \epsilon < a < a + \epsilon$ and $\displaystyle b- \epsilon < b < b+ \epsilon$. Then $\displaystyle 0 \leq b_n- a_n < b-a$ for all $\displaystyle n \in \mathbb{N}$. It follows that $\displaystyle a \leq b$. $\displaystyle \square$

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2. Originally Posted by manjohn12
Prove the following: Let $\displaystyle (a_n)$ and $\displaystyle (b_n)$ be sequences of real numbers that converge to real numbers $\displaystyle a$ and $\displaystyle b$ respectively. If $\displaystyle a_n \leq b_n$ for all $\displaystyle n \in \mathbb{N}$, then $\displaystyle a \leq b$.

Proof: For each positive real number $\displaystyle \epsilon$ there exists a $\displaystyle N_1 \in \mathbb{N}$ so that whenever $\displaystyle n > N_1$, $\displaystyle |a_n-a| < \epsilon$. Also for each positive real number $\displaystyle \epsilon$ there exists a $\displaystyle N_2 \in \mathbb{N}$ so that whenever $\displaystyle n > N_2$, $\displaystyle |b_n-b| < \epsilon$. So $\displaystyle a- \epsilon < a < a + \epsilon$ and $\displaystyle b- \epsilon < b < b+ \epsilon$. Then $\displaystyle 0 \leq b_n- a_n < b-a$ for all $\displaystyle n \in \mathbb{N}$. It follows that $\displaystyle a \leq b$. $\displaystyle \square$

Is this correct?
Try proving the following easier result: let $\displaystyle \{ x_n\}$ be a sequence such that $\displaystyle x_n \geq 0$ and $\displaystyle x_n \to x$ then $\displaystyle x\geq 0$.

After you prove that consider $\displaystyle x_n = a_n - b_n \geq 0$.